Bresenham’s Line Algorithm

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The Problem (cont…)

What happens when we try to draw this on a pixel based display?

How do we choose which pixels to turn on?

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In Bresenham’s line drawing algorithm the incremental integer calculations are used to scan convert the lines, so that, the circles and the other curves can be displayed.

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.The next sample positions can be plotted either at (3,2) or (3,3) ?

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(2,2) (3,2)

(3,3)

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For lines with positive slope m<1The pixel positions on line can be identified by doing sampling at unit x intervals. The process of sampling begins from the pixel position (X0,Y0) and proceeds by plotting the pixels whose ‘Y’ value is nearest to the line path. If the pixel to be displayed occurs at a position (Xk, Yk) then the next pixel is either at (Xk+1,Yk) or (Xk+1,Yk+1) i.e, (3,2) or (3,3) The ‘Y’ coordinate at the pixel position Xk+1 can be obtained from

Y=m(Xk+1)+b . . . . . . . . . . ……. eq 1

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the separation between (Xk+1,Yk) and (Xk+1,Y) is d1 and the separation between (Xk+1,Y) and (Xk+1, Yk+1) is d2 then

d1 = y – yk and

d2 = (Yk+1) – Y .

(Xk, yk)

(Xk, yk+1) (Xk+1, yk+1)

(Xk+1,yk)

P0 d1

d2

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Y=m(Xk+1)+b . . . . . . . ……. eq 1d1=y – yk

d1=m(xk+1)+b – Yk ( from eqn (1) …..( 2) And d2= (Yk+1) – Y

=(YK+1) – [m(Xk+1)+b]=(YK+1) – m(Xk+1) – b …….(3)

The difference is given asd1-d2 =

= m(Xk+1)+b-Yk-[(Yk+1)-m(Xk+1)-b] =m(Xk+1)+b-Yk-(Yk+1)+m(Xk+1)+b

d1-d2 = 2m(Xk+1)-2Yk+2b – 1 ………(4)

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Contd..A decision parameter Pk can be obtained by substituting m= dy/dx in equation 4

d1 – d2 = 2m(Xk+1)-2Yk+2b – 1 = 2 dy/dx (Xk+1) – 2Yk + 2b – 1 = 2 dy(Xk+1)-2 dx.Yk + 2b.dx -dx

dxdx(d1-d2) = 2 dy(Xk+1)-2 dx.Yk + 2b.dx - dx

= 2 dyXk+2 dy-2 dx.Yk + 2b.dx -dx

= 2 dyXk- 2 dx.Yk + c Where, dx(d1-d2) = Pk and

c= 2 dy+ dx(2b-1)Pk = 2 dyXk- 2 dx.Yk + c . . . . . . . . . . . . . . . . . . . … …(5 )

The value of c is constant and is independent of the pixel position. It can be deleted in the recursive calculations, of for Pk

if d1 < d2 (i.e, Yk is nearer to the line path than Yk+1) then, Pk is negative. If Pk is –ve, a lower pixel (Yk)is plotted else, an upper pixel (Yk+1)is plotted.

At k+1 step, the value of Pk is given asPK+1 = 2 dyXk+1- 2 dx.Yk+1 + c …………………………………..(6 ) (from 5)

The starting parameter P0 at the pixel position (X0,Y0) is given as P0 = 2dy – dx

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(X0, y0)

(X0, y0+1) (X0+1, y0+1)

(X0+1,y0)

P0 d1

d2

y=mx+c is the eq of lineIn col 2 the line is passing through x0+1 so the y value is given byy=m(x0+1)+c (green dot)Now we need to find out the values of d1 and d2

d1= y-y0

d2=(y0+1)-yd1=m(x0+1) – y0 and d2= (y0+1) - m(x0+1)

d1-d2=[mx0+m – y0] - [(y0+1) - mx0-m] =mx0+m – y0 - y0-1 + mx0+m = 2mx0+2m –2y0-1 = 2mx0+2m –2mx0-1 (y=mx+c passes thr (x0, y0) so we can say y0=mx0+c)d1-d2 = 2mx0+2m –2mx0-1 d1-d2=2m-1 ( m= ∆y / ∆x)d1-d2 = 2∆y/ ∆x - 1

P0= 2∆y - ∆x

Col 1 Col 2

∆x(d1-d2) = 2∆y - ∆x ∆x(d1-d2) = 2∆y - ∆x

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Bresenham’s algorithm

Step 1: Enter the 2 end points for a line and store the left end point in (X0,Y0).

Step 2: Plot the first point be loading (X0,Y0) in the frame buffer.Setp 3: determine the initial value of the decision parameter by

calculating the constants dx, dy, 2dy and 2dy-2dx asP0 = 2dy –dx

Step 4: for each Xk, conduct the following test, starting from k= 0

If Pk <0, then the next point to be plotted is at (Xk+1, Yk) andPk+1 = Pk + 2dy

Else, the next point is (Xk+1, Yk+1) and Pk+1 = Pk + 2dy –2dx (step 3)

Step 5: iterate through step (4) dx times.

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Task

Let the given end points for the line be (30,20) and (40, 28)

Track the line using Bresenham Algortithm.

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The starting point (x0, y0)=(30,20) and the successive pixel positions are given in the following table

K Pk (Xk+1, Yk+1)

0 6 (31,21)1 2 (32,222 -2 33,223 14 34,234 10 35,245 6 36,256 2 37,267 -2 38,268 14 39,279 10 40,28

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End points (30,20) (40,28)dx=x2-x1=40-30=10 dy= y2-y1=28-20=8P0=2dy-dx=2X8-10 = 6 >0 pt(31,21)Pk+1= pk+2dy-2dx= 6+16-20 =2 >0 (32,22)Pk+1=2+16-20 =-2 <0 (33,22)Pk+1= pk+2dy =-2+16 = 14 >0 (34,23)Pk+1= pk+2dy-2dx=14+16-20=10>0 (35,24)Pk+1=10+16-20=6 >0 (36,25)Pk+1= 6+16-20=2>0 (37,26)Pk+1=2+16-20=-2 <0 (38,26)Pk+1= pk+2dy=-2+16=14>0 (39,27)Pk+1=pk+2dy-2dx= 14+16-20=10>0 (40,28)

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Advantages of Bresenham’s Alg.

It uses only integer calculations So, it is faster than DDA